Aim laser canon on an incline
This is a story problem related to a more complex real problem. I am an engineer and often solve problems similar to this one, but this one so far has me beat. So I will appreciate your help if you can do it.(Thinking)
I have a tiny army tank with a laser cannon in a large room with a 35 foot ceiling. The tank sits at the origin x,y,z=0, and can hit any target point on the ceiling given the T(x,y,35) coordinates based on the following formulas for turret rotation angle(Ar) and canon elevation angle(Ae):
Ae= ArcTan(35/(SqRt(x^2 + y^2))
Then if due to an explosion, the tank is left in position at (0,0,0) but angled 15 degrees upward on a plane rotated about the x-axis, what are the new formulas using the same x,y,z for the ceiling, but angles relative to the new plane of the tank, Aen and Arn ("n" for "new").
1. Ar,Arn = 0 when the tank turret is parallel to the tanks x,z(or zn) plane. Range is -180 to +180 deg.
2. Ae = 0 when the tanks turret is parallel to the z-axis(or zn-axis). Range is from -90 to +90.